JEE PYQ: Application Of Derivatives Question 23
Question 23 - 2021 (26 Feb Shift 2)
Let the normals at all the points on a given curve pass through a fixed point $(a, b)$. If the curve passes through $(3, -3)$ and $(4, -2\sqrt{2})$, and given that $a - 2\sqrt{2},b = 3$, then $(a^2 + b^2 + ab)$ is equal to
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Answer: 9
Solution
Normals pass through $(a,b)$: the curve is a circle with center $(a,b)$. The curve passes through $(3,-3)$ and $(4,-2\sqrt{2})$. Using the distance conditions and $a - 2\sqrt{2},b = 3$: $a = 3, b = 0$. So $a^2 + b^2 + ab = 9$.
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